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Homework Help: Simple Harmonic Motion, platform and box

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A platform moves up and down in SHM, with amplitude 0.035 m. Resting on top of the platform is a block of wood. What is the shortest period of motion for the platform so that the block will remain in constant contact with it?


    2. Relevant equations

    a(t)=-A[tex]\omega[/tex]^2cos([tex]\omega[/tex]t+phase constant)
    amax=A[tex]\omega[/tex]

    3. The attempt at a solution

    I didn't see how I could possibly use the first one with so many unknowns so I used amax=A[tex]\omega[/tex] and set amax=9.8 figuring that the amount couldn't be more than gravity otherwise the block and platform would separate (maybe I'm wrong in this).
    And so I set 9.8=.035[tex]\omega[/tex] and solved (getting 280). The I used [tex]\omega[/tex]=2[tex]\pi[/tex]f and solving for f (getting 140/[tex]\pi[/tex]) and then using f=1/T and solved for T getting .0224 which was incorrect.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 6, 2010 #2

    ehild

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    Homework Helper

    Check your formula for amax, it is not correct.

    ehild
     
  4. Apr 6, 2010 #3
    Oh, it came out funny. It's supposed to be amax=A x omega. Is this still incorrect? If it is, I really need to throw a fit about a refund.
     
  5. Apr 6, 2010 #4

    ehild

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    Homework Helper

    No, see the previous line. a must be m/s2, yours is m/s.

    ehild
     
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